General Introduction
Definition of decision: The act of process of choosing one course of action from among several
alternatives.
Alternative defintion: The act of selecting one option from among two or more possibilities.
There are four different ways of approaching the problem of choice.
These are the Inactive, Reactive, Proactive and Interactive approach.
1. Inactive approach: The decision maker does nothing of consequences to make a thoughtful
choice. In effect the choice is to let the problem resolve itself. This approach is quite common in
everyday decision making. This can not be considered rational because we mostly did not do anything
to favorably influnce the outcome.
2. Reactive approach: The decision maker opts for a course of action by reacting to the choices
made by other stakeholders of the decision problem. This approach is also quite common in our
everyday life and in ordinary decision making. Be careful !!! reacting to the actions of others without
first analyzing the overall situation can easily expose the decision maker to such situations what can
not be resoled after that. This approach can easily lead our business or enterprises into unforeseen
3. Proactive approach: The decision maker opts for a course of action by FIRST CAREFULLY
ANALYZING the problem situation within its relevant context (the environment in which the problem
situation is embedded). This is the normative approach to decision making in professional
management.
4. Interactive approach: The decision maker probes (if it is possible !!!) the underlying problem by
experimenting with tentative courses of action before making the final choice. In effect the decision
problem itself is used as a test bed to explore the feasibility and desirability of possible alternative
solutions.
Rational decision making process:
1. Recognize and define the problem: This is the process of correctly identifying the actual problem
or opportunity faced by the decision maker.
2. Gather information: Information gathering means to collect relevant facts related to the decision
problem. This often reduces to a search problem.
3. Identification of action alternatives: As the decision maker gathers information the decision maker
begins crystallizing possible solution alternatives. The emphasis at this point should be on generating
possible courses of action, not on criticizing or evaluating the alternatives.
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4. Determine the evaluation criteria: In order to evaluate the action alternatives the decision maker
must determine the evaluation criteria and the relative importance of each criterion. The evalution
criteria is mostly a some kind of quatitative measure.
5. Evaluate the alternatives: The decision maker compares the pors and cons of each action
alternative in accordance with the evaluation criteria.
6. Select the best alternative: This is the classic decision making point.
7. Implement the chosen alternative: The decision maker sets in motion a course of action that
involves the customary managerail tasks.
8. Evaluate the results: This could be a learning process from our successes or mistakes.
Tools of decision making to improve the managerial decisions.
These are mostly some kind of mathematical tools or modeling to effectively evaluate the arising
problem or opportunity.
1. Linear programming
2. Stochasting modeling of decision problems
3. Network evaluation (like Pert Network) critical path
Linear programming
This is a very efficient mathematical tool to optimally use or distribute resources between the arising
needs. Historically it has been developed under the second world war in Great Britain when was under
siege by the submarine war between 1939-1945.
Programming means some kind of manufacturing program where we produce or create some amounts
of products.
Basic part of a Linear programming model.
1. Decision variables
2. Object function (this can be transformed to a linear equation)
3. Contraints (this is the linear inequality system)
4. Auxiliary data
Case Study or Sample Task
The Task: There is Mr. Tichmaher a carpenter (making furniture) manufacturing tables and chairs. He
has 16 + 2/3 wood plate, 600 dm wood stick and 50 working days. Manufacturing a table it needs 1
wood plate, 25 dm wood stick, 1 working day and producing a chair it needs 1/3 wood plate, 15 dm
wood stcik, 2 working days. Mr. Tichmaher can sell a table with 300 Eu/table and a chair with unit
price 150 Eu/chair.
The question: How much tables and chairs have to be manufactured to maximize the income and how
much is that (income)?
First step is to develop a mathematical model
Second step is to choose the most efffective method to solve the mathematical model. This could be
Analytic (using simplex method for example), Grapho-Analitic (only usede with two decision
variables), some kind of computer simulation tools (most simplest is the Excel/Solver)
Third step is to Solve the problem one of the above mentioned tools.
First thing: Develop the mathematical model.
1. Decision variables: How much is that? How much decision variables we have?
The carpenter are manufacturing two different kind of products (table, chair).
The number of tables indicated by x, The number of chairs indicated by y. And unit (dimension)
of these products (the manufactured number of tables and chairs)
2. Object function: How can we formulate the income function?
(300 𝑥 + 150 𝑦)
What can we do with this? (Maximize, Minimize or set a specified number)
𝑚𝑎𝑥 (300 𝑥 + 150 𝑦)
3. Constraints: Our resources allways limited!!! The constraints try to describe this FACT
mathematically.
The general form of contraints looks like this: (Sentence-like description of a mathematical formula)
The used or consumed resources <= or = Available resources 25 𝑥 + 15 𝑦 ≤ 600 𝑥 + 2 𝑦 ≤ 50 How can we solve this mathematical model? 1. Simplex method 2. Graphical method (because two dimencsional we have two decision variables) 3. Can be solved by some kind of computer program developed to solve LP problems. What kind of mathematical tools are needed to learn the Simplex method? The answer is: Vector algebra (linear algebra) and some knowledge of linear space. 1. Concept of linear combination of vectors, linear independece of vectors 2. Concept of the base, base vectors and the meaning and techniques of base transformation. 3. Concept of rank of a vector system and matrices. 4. Solution of linear equation systems 5. Simplex method (application) 6. Disributional problems (transportation problems) (application) https://www.superprof.co.uk/resources/academic/maths/linear-algebra/linear-programming/linear- programming-examples.html https://www.superprof.co.uk/resources/academic/maths/linear-algebra/linear-programming/linear-programming-examples.html https://www.superprof.co.uk/resources/academic/maths/linear-algebra/linear-programming/linear-programming-examples.html Concept of the linear combination. Definition: Let suppose we have a linear space (I do not defined what is this! In the following this will be a vector space but it is importatnt to note that for example functions (polinomials) can also be create a linear space) with containing the following n piece of vectors. 𝑣1 ; 𝑣2 ; … ; 𝑣𝑛 We say the following arrangement of vectors 𝑐1 𝑣1 + 𝑐2 𝑣2 + ⋯+ 𝑐𝑛 𝑣𝑛 is the linear combination of the vectors 𝑣1 ; 𝑣2 ; … ; 𝑣𝑛 . Where 𝑐1 , 𝑐2 , … , 𝑐𝑛 are scalar values. We use two basic operations on vectors: The first one is addition of vectors 𝑣1 + 𝑣2 and the second one is the scalar multiplication 𝑐1 𝑣1. Some simple example for these operations because of reminding. wvu . The addition of 𝐮 + 𝐯 =[ ] and a scalar multiplication can be Example of a linear combination: 2 𝐮 − 𝐯 + 𝐰 = 2 [ Some kind of visualisation of vectors in three dimensions: Let suppose we have the following three vectors creating 3 dimensional linear vector space. e1=[1, 0, 0] T, e2=[0, 1, 0] T, e3=[0, 0, 1] T. In addition to this we have a vector v=(a, 0, b). This can be created with the following linear combination of vector [e1, e2, e3]. Concept of linear indepence of vectors. (Definition) Let suppose we have a linear space containing the following n piece of vectors 𝑣1 ; 𝑣2 ; … ; 𝑣𝑛 and we have 𝑐1 , 𝑐2 , … , 𝑐𝑛 scalar values. We say these vectors are linearly independent if and only if that the following linear combination 𝑐1 𝑣1 + 𝑐2 𝑣2 + ⋯+ 𝑐𝑛 𝑣𝑛=0 can (create) forms the null vector if all values 𝑐1 , 𝑐2 , … , 𝑐𝑛 scalars are zero. This is slightly a complicated idea (or concept). Because this is not directly formulated with the real meaning hidden behind that. This is a little bit indirect description of the matter (essence) of meaning. Some notation about the symbols: 1. This means a scalar value: x, y, a, b, … 2. This means a vector: 𝑣1 , v. 3. This means a matrix: A, B The meaning of independence is allways arising between a group of vectors!!!! Important!!! If they are independent to each other this means that can not be produced (formulated with a linear combination) any of them with the others. Arising the question: Is it possible to create the null vector 0 with the following linear combination of vectors 𝑐1 𝑣1 + 𝑐2 𝑣2 + ⋯+ 𝑐𝑛 𝑣𝑛 ? and If it is possible in that case how??? Of course in a trivial way c1=0, c2=0, …, cn=0 it is natually possible. BUT!!! Are there no zero values of 𝑐1 , 𝑐2 , … , 𝑐𝑛 to create the null vector? The answer is: Sometimes YES and sometimes NO, depending on the vectors 𝑣1 ; 𝑣2 ; … ; 𝑣𝑛 . Example 1. Investigate the following three vectors wvu that what kind of linear combination of vectors u, v, w can create the null vector? Solution: We have to check and solve the following linear equation system, c1 u + c2 v + c3 w = 0 (1.) because we would like to know the values of c1, c2, c3. If we write out the the vector equation (1.) in a more detailed form And after that We expand the vector equation (1.) in the following scalar form We get the following linear equation system Let solve it! Currently I use a secondray school techique to solve the equation system above. Because second equation does not contain the unknown coefficient c2 for this reason we try to eliminate this from the first and third one also. We substract the first equation (I.) from two times of the third equation (III.) with the following way 2 III. – I and get the following two equations. 5 c1 + 15 c3 =0 3 c1 + 9 c3 =0. Express c1 from any of the above equations we get c1 = – 3 c3. From this the general solution of the vector equation (1.) is the following: c1 = – 3 c3 , c2 = – c3. THE CONCLUSION: We have infinite number of possible values of c1, c2, c3 to produce the null vector from vectors wvu . For example let be c3 = 6, c1 = -18, c2 = -6, Let’s check that is that really produce null vector or not!! Is it equal or not?? -18 u - 6 v +6 w ?? = ?? 0 We substitute the general solution c1 = – 3 c3 , c2 = – c3 into the vector equation (1.) c1 u + c2 v + c3 w = 0 and got – 3 c3 u – c3 v + c3 w = 0 we simplify this with c3 and we got w = 3 u + v . We can say that these vectors u, v, w are linearly dependent. Example 2. Prove that the following three vectors wvu can only create the null vector with the trivial way. Solution: We have to check and solve the following vector equation c1 u + c2 v + c3 w = 0 (2.) because we would like to know the values of c1, c2, c3. If we write out the the vector equation (2.) in a more detailed expanded form We can reformulate it in a linear equation system what has the following form This is a homogeneous equation system because all values on the right side of the equation are zero. From the second equation we imideately see that c2=0 and if it is in that case from the third equation we see that c3=0 and finally from the first equation we can conclude that the first coefficient is also zero c1=0. The solution is c1 = 0, c2 = 0, c3 = 0. THE CONCLUSION: We have only one possible combination values of c1, c2, c3 to produce the null vector from vectors wvu . These values are zeros and this can be called the trivial linear combination of vectors u, v, w . So ONLY the following combination (0 u + 0 v + 0 w = 0) of vectors u, v, w can produce the null vector. The most important outcome of the above result is that we CAN NOT express any of these vectors u, v, w from the remaining others. Other words there is not exist a linear combination of u ≠ v + w or v ≠ u + w or w ≠ v + u where and are arbitrary real scalar numbers. So we can say the vectors u, v, w are linearly independet! Concept of the base of a vector system. Concept of BASE Definition: We can say that a set of vectors is a base if the vectors are linearly independent and can produce (create) all other vectors uniquely in the vector space with the appropriate linear combination. In a vector space generally there are many (or infinite number of) different base, that means we can change the different base in a vector space. Some example for the base. 1. Example: Three dimensional elementary base. The base vectors are the following ]1,0,0[,]0,1,0[,]0,0,1[ 321 eee and the geometrical interpretation can be seen below. If we would like to create a vector using the base vectors we have to apply the following linear combination, v = v1 e1 + v2 e2 + v3 e3 2. Example: ]3,0[,]0,2[ 21 vv Orthogonal base in x-y plane 3. Example: n dimensional trivial base: The base vectors are the following: ]1,,0,0[,,]0,,1,0[,]0,,0,1[ eee And an arbitrary vector can be created with the following linear combination: 𝑎1 𝐞1 + 𝑎2 𝐞2 + ⋯+ 𝑎𝑛 𝐞𝑛=𝐚 in the n dimensional vector space. Concept of the base transformation: This is the first important ingredient solving the Case study problem (LP problem) presented in the first lecture. General description: First let be the n dimensional trivial base is indicated by the following set of vectors [𝐞1 ; 𝐞2 ; … ; 𝐞𝑛 ] and this will be indicated shortly by the following matrix E. In this base we have a vector x created by the following linear combination: 𝑥1 𝐞1 + 𝑥2 𝐞2 + ⋯+ 𝑥𝑛 𝐞𝑛=𝐱. If we have an another base indicated by V what has the following set of vectors [𝐯1 ; 𝐯2 ; … ; 𝐯𝑛 ] and the same vector x can be created with the following linear combination: 𝑐1 𝐯1 + 𝑐2 𝐯2 + ⋯+ 𝑐𝑛 𝐯𝑛=𝐱 Arising the following question: If we know the scalar components (𝑥1 , 𝑥2 , … , 𝑥𝑛 ) of vector x in the original (or trivial) base E and we know the new base vectors [𝐯1 ; 𝐯2 ; … ; 𝐯𝑛 ] how can we calculate the unknown components of vector x in the new base? The following general approach can be followed. In matrix notation veector x can be produced by the following two different ways: x = E xE = V xV (1.) where xE indicate the scalar components of vector x in the trivial base and xV indicate the scalar components of vector x in the new base. Let multiply both side of matrix equation (1.) with inverse of matrix V . The inverse is indicated V-1 E xE = V Applying the dfinition of the inverse matrix that V-1 V = E it is equal with the identity matrix E and V-1 E = V-1 we get the following final result. V-1 xE = E xV and finally xV = V So the general receipt (or prescription) to calculate the scalar components of a vector in a new base is to calculate the inverse of matrix formed by the new base vectors. General notation: Calculating matrix inverses is a very difficult and computationally cost ineffective task. We can resolve this problem that only calculate such steps what are needed to us. So we can not calculate the entire inverse instead of it we calculate the so called elementary base transformations. But before this I would like to show you a simple example of the above detailed concept. Example: Let specify the following two linearly independent vectors a new base in the x-y plane vv and a vector x has the following scalar components Ex in the elementary base e1, e2, in other words 21 45 eexEx E Calculate the new scalar components of vector x in the new base indicated by V. Indication vectors in the above task, coordinats of xE vector in the base e1, e2 base and the new base (v1, v2) We solve this with two different ways: 1. The first is to use the so called Gauss-Jordan method to calculate the inverse of the matrix of the new base vectors. 2. Step by step with elementary base transformation calculate the same result. This approach is somewhat a secondary school technique. Solution with the Gauss-Jordan method: First create the matrix V from the new base vectors. ] the task to calculate V-1 =? . We start with the following arrangement of matrix V and the identity matrix E [ Applying (using) the inverse matrix we can calculate the new scalar components of vector x in the new base. Vxvvx 21 Let’s calculate this a more elementary way step by step. The task is to calculate the new scalar components of vector x in three different base {(v1; e2); (e1; v2); (v1; v2)}. First we write (v1 ; v2) vectors in the trivial base with the following algebraic form. v1 = 3 e1 + 2 e2; v2 = 2 e1 + 4 e2. xE vector coordinates in base e1, e2 and the new base vectors (v1, v2) First if we choose (v1; e2) to a base then we have to express vector e1 with vectors (v1; e2) . We get the following v1 = 3 e1 + 2 e2; 211 eve , after that substitute into the vector equation x = 5 e1 + 4 e2, describing vector x and we get vector x in the base (v1; e2) xE vector in the new base v1, e2 If we choose an other base (e1; v2) in that case we express vector e2 with vectors (e1; v2). The result is 122 eve , and after that we substitute into the expression of vector x = 5 e1 + 4 e2 and we have got in the new base (e1; v2) vector x 213 vex . The next figure shows this transformation. Components of vector xE in the new base e1, v2 Finally if we choose (v1; v2) to the new base vectors, we have to solve this in two steps. First one of the original unit base vectors (e1; e2) has to expressed one of the vectors (v1; v2) and after this result has to be substituted into the equation of vector x. For example express vector e2 with vectors (e1; v2) 122 eve , after this substitute into the expression of vector x , x = 5 e1 + 4 e2, and substitute this into the expression of vector v1 v1 = 3 e1 + 2 e2. We have got the following two expressions in the new base (e1; v2) for vectors x and v1. After this express vector e1 from the expression of 211 2 vev 211 and after this step substitute this expression 211 vve into 213 vex and we have got vector x in the new base (v1; v2). Compare this with the result has been got by the method of Gauss, Jordan. Components of vector xE in the new base described by vectors v1, v2 . 5.2. Generalization of the elementary base transformation in arbitrary dimensions (scalar components of a vector). Let’s consider the n dimensional vectors space Vn. In this vector space vectors b1; b2; ..., bn; indicate n linearly independent base vectors what can form a new base. Let be vector c specified by the following form c = c1 b1 + c2 b2 + ... + ci bi + ... + cn bn. (1.) cccc ,,,,, real scalar numbers are the coordinates of vector c in the base represented by vectors b1; b2; ..., bi; ..., bn;. Vector c can be specified in matrix notaion this way cccc ,,,,, c . Let be vector x specified in the following form x = x1 b1 + x2 b2 + ... + xi bi + ... + xn bn, (2.) xxxx ,,,,, real numbers are the scalar components (so called coordinates) of vector x in the base b1; b2; ..., bi; ..., bn; what means in matrix notation xxxx ,,,,, We have to specify (calculate) the coordinates of vector x or express vector x with the following base vectors b1; b2; ..., c; ..., bn;. You might have noticed that we replace vector bi with vector First step is to express vector bi from the vector equation (1.) of vector c. Rearrange: – c1 b1 – c2 b2 – ... + c – ... – cn bn = ci bi. (1.) After some simple elementary algebra we got bbcbbbb Second step: We substitute of expression (3.) describing vector bi into the vector equation (2.) expressing vector x. With this step we eliminate vector bi from the the base b1; b2; ..., bi; ..., bn; and replace vector bi with vector c . After the necessary algebraic arrangements we got the fololwing expression for vector x . x bbcbbx It is practical if we summarize the above detailed calculations in a table. Old base c x New base bi x b1 c1 x1 b1 b2 c2 x2 b2 bi-1 ci-1 xi-1 bi-1 bi ci xi c bi+1 ci+1 xi+1 bi+1 bn cn xn bn 5.2.1. Example: Let be two four component vectors specified by the following matrix notation: v1=[1; 2; 5; 4] T, v2=[1; 5; 2; 2] T, These vectors are specified in the elementary trivial base. v1= e1 + 2 e2 + 5 e3 + 4 e4 and v2= e1 + 5 e2 + 2 e3 + 2 e4. The task is to calculate the coordinates of vector v1 in the base v2; e2; e3; e4 and the coordinates of vector v2 in the base v1; e2; e3; e4. Az alábbi táblázat a v1 vektor v2; e2; e3; e4 bázisbeli koordinátáinak meghatározását mutatja. v1 v2 v1 e1 v1 e1 e2 2 5 e2 2 - e3 5 2 e3 5 - e4 4 2 e4 4 - v1= v2 – 3 e2 + 3 e3 + 2 e4 e1= v2 – 5 e2 – 2 e3 – 2 e4 Generator element: it is always a denominator in the calculations Concept of rank. Definition: In linear algebra the RANK means the maximum number of linearly independent elements of a linear system. Equivalently in a vector system: The rank of a vector system is the maximum number of linearly independent base vectors. Example 4.1.1. A vector systems contains the following vectors v1=[1; 0; 2; 4] T, v2=[0; 5; 2; 1] v3=[1; 25; 12; 9] T, v4=[1; -5; 0; 3] T, v5=[1; 10; 6; 6] T. Calculate the rank of the vectors system specified by the vectors seen above. This equivalently means that how much vectors can be carried into the base? We have to count the number of the new base vectors. v1 v2 v3 v4 v5 v2 v3 v4 v5 e1 1 0 1 1 1 v1 0 e2 0 5 25 -5 10 e2 5- 0=25 -5 - 0 =-5 10 – e3 2 2 12 0 6 e3 2- 2 =10 0 - e4 4 1 9 3 6 e4 1-- v2 v3 v4 v5 v3 v4 v5 v1 0 1 1 1 v1 1- e2 5 25 -5 10 e2 25- 5 = 0 10 – e3 2 10 -2 4 v2 e4 1 5 -1 2 e4 5- The conclusion: How much is the rank? The rank is 2 because we can put two vectors into the How can we formulate the vectors (v3, v4, v5) remain outside the base? v3 = v1 + 5 v2 v4 = v1 - v2 v5 = v1 + 2 v2 Definition: We say a vector u is compatible with a vector system describing with the following vectors v1; v2; ..., vn; if vector u can be expressed (or calculated) by the linear combination of vectors v1; v2; ..., vn;. Example 4.2.1 A vector systems contains the following vectors v1=[1; -1; 2; 3] T, v2=[2; 0; 4; 1] T, v3=[4; -2; 8; 7] T, v4=[-3; -1; -6; 1] T, v5=[-1; 3; -2; -8] additionally we have two other vectors u=[9; -5; 18; 17]T, w=[-1; -1; -2; 4]T. Investigate that vectors u and w are compatible or not with the vector system specified by v1; v2; ..., v5;. v1 v2 v3 v4 v5 u w v2 v3 v4 v5 u w e1 1 2 4 -3 -1 9 -1 v1 2 4 -3 -1 9 -1 e2 -1 0 -2 -1 3 -5 -1 e2 2 2 -4 2 4 -2 e3 2 4 8 -6 -2 18 -2 e3 0 0 0 0 0 0 e4 3 1 7 1 -8 17 4 e4 -5 -5 10 -5 -10 7 The next table details the calculations to continue the solution: v2 v3 v4 v5 u w v1 2 4 -3 -1 9 -1 e2 2 2 -4 2 4 -2 e3 0 0 0 0 0 0 e4 -5 -5 10 -5 -10 7 v3 v4 v5 u w e3 0 0 0 0 0 (-5)=0 10- (-5)=0 -5- (-5)=0 -10- Evaluate the content of the table above: Vectors v3; v4; v5; and vectors u and w can be expressed in the new base with the following v3 =2 v1 + v2 v4 = v1 -2 v2 v5 =-3 v1 + v2 u =5 v1 +2 v2 w = v1 - v2 +2 e4 What is the conclusion? From the results we can see that vector u is compatible with the vector system v1; v2; ..., v5; because it can be expressed by a linear combination of these vectors. But otherwise vector w is not compatible with the vector system v1; v2; ..., v5; because it can not be expressed with these vectors. To express vector w vector e4 is necessary. Practice of base transformation Practicing example: Make a decision that the vectors u=[16; -19; 11; 10]T, w=[1; -8; 8; -6]T are compatible or not with the following vector system v1=[6; -4; 4; -6] T, v2=[1; 1; 0; 2] T, v3=[7; -10; 5; 8] v4=[5; -4; 3; 2] T, v5=[3; -5; 2; 8] v1 v2 v3 v4 v5 u w v1 v3 v4 v5 u w e1 6 1 7 5 3 16 1 v2 6 7 5 e2 -4 1 -10 -4 -5 -19 -8 e2 -10 -17 -9 -8 -35 -9 e3 4 0 5 3 2 11 8 e3 4 5 3 2 11 8 e4 -6 2 8 2 8 10 -6 e4 -18 -6 -8 2 -22 -8 v1 v3 v4 u w v1 v4 u w v2 33 16 17 49 13 v2 1 1 1 - -82 -41 -41 -123 -41 22 11 11 33